Selina solutions

Selina solutions

Grade 7

Integers | Exercise 1(D)

Question 11

Q11) Evaluate :

(i) \left(-20\right)+\left(-8\right)\div\left(-2\right)\times3

(ii) \left(-5\right)-\left(-48\right)\div\left(-16\right)+\left(-2\right)\times6

(iii) 16+8\div4-2\times3

(iv) 16\div8\times4-2\times3

(v) 27-\left[5+\left\{28-\left(29-7\right)\right\}\right]

(vi) 48-\left[18-\left\{16-\left(5-\overline{4+1}\right)\right\}\right]

(vii) -8-\left\{-6\left(9-11\right)+18\div\left(-3\right)\right\}

(viii) \left(24\div\overline{12-9}-12\right)-\left(3\times8\div4+1\right)

Solution 11:

(i) \left(-20\right)+\left(-8\right)\div\left(-2\right)\times3

= -20+\frac{-8}{-2}\times3

= -20+4\times3

= -20+12

= -8

(ii) \left(-5\right)-\left(-48\right)\div\left(-16\right)+\left(-2\right)\times6

= \left(-5\right)-\frac{-48}{-16}+\left(-2\right)\times6

= \left(-5\right)-3+\left(-12\right)

= -5-3-12

= -20

(iii) 16+8\div4-2\times3

= 16+\frac{8}{4}-2\times3

= 16+2-6

= 18-6=12

(iv) 16\div8\times4-2\times3

= \frac{16}{8}\times4-2\times3

= 2\times4-6

= 8-6=2

(v) 27-\left[5+\left\{28-\left(29-7\right)\right\}\right]

= 27-\left[5+\left\{28-22\right\}\right]

= 27-\left[5+6\right]

= 27-11=16

(vi) 48-\left[18-\left\{16-\left(5-\overline{4+1}\right)\right\}\right]

= 48-\left[18-\left\{16-\left(5-5\right)\right\}\right]

= 48-\left[18-\left\{16-0\right\}\right]

= 48-\left[18-16\right]

= 48-2=46

(vii) -8-\left\{-6\left(9-11\right)+18\div\left(-3\right)\right\}

= -8-\left\{-6\left(-2\right)+\frac{18}{-3}\right\}

= -8-\left\{12+\left(-6\right)\right\}

= -8-6=-14

(viii) \left(24\div\overline{12-9}-12\right)-\left(3\times8\div4+1\right)

= \left(24\div3-12\right)-\left(3\times\frac{8}{4}+1\right)

= \left(\frac{24}{3}-12\right)-\left(3\times2+1\right)

= \left(8-12\right)-\left(6+1\right)

= -4-7=-11

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Question 11

Q11) Evaluate :

(i) \left(-20\right)+\left(-8\right)\div\left(-2\right)\times3

(ii) \left(-5\right)-\left(-48\right)\div\left(-16\right)+\left(-2\right)\times6

(iii) 16+8\div4-2\times3

(iv) 16\div8\times4-2\times3

(v) 27-\left[5+\left\{28-\left(29-7\right)\right\}\right]

(vi) 48-\left[18-\left\{16-\left(5-\overline{4+1}\right)\right\}\right]

(vii) -8-\left\{-6\left(9-11\right)+18\div\left(-3\right)\right\}

(viii) \left(24\div\overline{12-9}-12\right)-\left(3\times8\div4+1\right)

Solution 11:

(i) \left(-20\right)+\left(-8\right)\div\left(-2\right)\times3

= -20+\frac{-8}{-2}\times3

= -20+4\times3

= -20+12

= -8

(ii) \left(-5\right)-\left(-48\right)\div\left(-16\right)+\left(-2\right)\times6

= \left(-5\right)-\frac{-48}{-16}+\left(-2\right)\times6

= \left(-5\right)-3+\left(-12\right)

= -5-3-12

= -20

(iii) 16+8\div4-2\times3

= 16+\frac{8}{4}-2\times3

= 16+2-6

= 18-6=12

(iv) 16\div8\times4-2\times3

= \frac{16}{8}\times4-2\times3

= 2\times4-6

= 8-6=2

(v) 27-\left[5+\left\{28-\left(29-7\right)\right\}\right]

= 27-\left[5+\left\{28-22\right\}\right]

= 27-\left[5+6\right]

= 27-11=16

(vi) 48-\left[18-\left\{16-\left(5-\overline{4+1}\right)\right\}\right]

= 48-\left[18-\left\{16-\left(5-5\right)\right\}\right]

= 48-\left[18-\left\{16-0\right\}\right]

= 48-\left[18-16\right]

= 48-2=46

(vii) -8-\left\{-6\left(9-11\right)+18\div\left(-3\right)\right\}

= -8-\left\{-6\left(-2\right)+\frac{18}{-3}\right\}

= -8-\left\{12+\left(-6\right)\right\}

= -8-6=-14

(viii) \left(24\div\overline{12-9}-12\right)-\left(3\times8\div4+1\right)

= \left(24\div3-12\right)-\left(3\times\frac{8}{4}+1\right)

= \left(\frac{24}{3}-12\right)-\left(3\times2+1\right)

= \left(8-12\right)-\left(6+1\right)

= -4-7=-11

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